Đáp án:
\(0 \le x \le \dfrac{3}{2}\)
Giải thích các bước giải:
\(\begin{array}{l}
\left| {2x - 1} \right| + \left| {3x - 2} \right| \le x + 3\\
\to {\left( {2x - 1} \right)^2} + 2\left( {2x - 1} \right)\left( {3x - 2} \right) + {\left( {3x - 2} \right)^2} \le {x^2} + 6x + 9\left( {DK:x \ge - 3} \right)\\
\to 4{x^2} - 4x + 1 + 2\left( {6{x^2} - 7x + 2} \right) + 9{x^2} - 12x + 4 \le {x^2} + 6x + 9\\
\to 24{x^2} - 36x \le 0\\
\end{array}\)
Xét:
\(\begin{array}{l}
24{x^2} - 36x = 0\\
\to \left[ \begin{array}{l}
x = 0\\
x = \frac{3}{2}
\end{array} \right.
\end{array}\)
BXD:
x -∞ 0 3/2 +∞
f(x) + 0 - 0 +
KL: \(0 \le x \le \dfrac{3}{2}\)
